Limit of time integral of brownian motion

Question

Can someone help explain the following,

$$ \lim \limits_{t \to 0} \frac{1}{t} \int_0^t W_u\, du=\lim \limits_{t \to 0} \frac{W_0t}{t}=W_0=0\,? $$

Thanks!

Answer 1

The fundamental theorem of calculus states that

$$\lim_{t \to 0} \frac{1}{t} \int_0^t f(u) \, du = f(0)$$

for any continuous function $f$. Applying this to $f(t) := W_t$ yields the result.

Answer 2

L'hopital? The limit is a $\frac{0}{0}$ type, so if we let $f(t) = \int_0^t W_udu$

$$\lim_{t\to0} \dfrac{f(t)}{t} = \lim_{t\to0} \dfrac{f'(t)}{1}$$

using fundamental calculus theorem $$f(t) = \int_0^tW_udu \implies f'(t)=W_t$$

$$\therefore \lim_{t\to0} \dfrac{f'(t)}{1} = \lim_{t\to0}W_t = W_0$$